/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 27 Let \(\alpha=3+4 i+7 j+k\) and \... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 21: Problem 27

Let \(\alpha=3+4 i+7 j+k\) and \(\beta=2-3 i+j-k\) be quaternions. Calculate \(\alpha \beta\) and \(\alpha / \beta\).

Short Answer

Expert verified
Question: Calculate the product and division of the quaternions \(\alpha = 3 + 4i + 7j + k\) and \(\beta = 2 - 3i + j - k\). Answer: The product of the quaternions is \(\alpha \beta = -23 - 7i - 9j + 9k\), and the division of the quaternions is \(\alpha / \beta = -1 + \frac{17}{15}i + \frac{2}{5}k\).

Step by step solution

01

Review quaternion multiplication rules

Quaternions are an extension of complex numbers and consist of a scalar part and a vector part. The quaternion multiplication follows these rules: 1. Scalars multiply with scalars and vectors using the distributive property. 2. The product of two vector parts follows the rules of cross product. 3. \(i^2 = j^2 = k^2 = ijk = -1\)
02

Multiply \(\alpha\) and \(\beta\)

Using the rules from Step 1, we can multiply the quaternions: \(\alpha \beta = (3+4i+7j+k)(2-3i+j-k)\) Now, expand the product and combine the terms: \(\alpha \beta = 6 - 9i + 3j - k + 8i^2 - 12ij + 4ik + 14ij - 21i^2 + 7j^2 - k^2 + 2ik - 3k^2\)
03

Simplify the product

We need to apply the rules for simplifying quaternion products: \(\alpha \beta = 6 - 9i + 3j - k - 8 - 12ij + 4ik - 21 - 7 - k^2 + 2ik + 3k^2\) Combine the like terms: \(\alpha \beta = -23 + (-9+2)i+ (-12+3)j+(4-1+6)k\) Therefore, \(\alpha \beta = -23 - 7i - 9j + 9k\)
04

Find the inverse of \(\beta\)

The inverse of quaternion \(\beta\) can be found by dividing the conjugate of \(\beta\) by the squared norm of \(\beta\): \(\beta^{-1} = \frac{\text{Conj}(\beta)}{||\beta||^2}\) The conjugate of \(\beta\) is: \(\text{Conj}(\beta) = 2+3i-j+k\) The squared norm of \(\beta\) is: \(||\beta||^2 = 2^2+(-3)^2+1^2+(-1)^2=4+9+1+1=15\) So, the inverse of \(\beta\) is: \(\beta^{-1}=\frac{1}{15}(2+3i-j+k)\)
05

Calculate \(\alpha / \beta\)

We can now find the division \(\alpha / \beta\) by multiplying \(\alpha\) with \(\beta^{-1}\): \(\alpha / \beta = \alpha \cdot \beta^{-1} = (3+4i+7j+k) \cdot \frac{1}{15}(2+3i-j+k)\) Now, expand the product, apply the rules from Step 1 and simplify: \(\alpha / \beta = \frac{1}{15}(6+9i-3j+3k+8i^2-12ij+4ik+14ij-7j^2-21i^2+k^2+2ik-3k^2) \) \(\alpha / \beta= \frac{1}{15}(-15+17i-k+7k)\) Finally, \(\alpha / \beta = -1 + \frac{17}{15}i + \frac{6}{15}k\) Which can be written as: \(\alpha / \beta = -1 + \frac{17}{15}i + \frac{2}{5}k\)

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Quaternion Multiplication
Quaternions extend complex numbers by introducing three imaginary units: \( i, j, \) and \( k \). Like complex numbers, quaternions have a real part and imaginary parts, but they expand into three dimensions. This makes quaternion multiplication unique. Here's how it works:
  • Use the distributive property. Multiply each part of the first quaternion by each part of the second.
  • Combine like terms, considering that \( i^2 = j^2 = k^2 = ijk = -1 \).
  • Cross product rules apply to imaginary units when multiplying (e.g., \( ij = k \)).
When multiplying quaternions \( \alpha = 3 + 4i + 7j + k \) and \( \beta = 2 - 3i + j - k \), start by expanding: \[ \alpha \beta = (3+4i+7j+k)(2-3i+j-k) \] Expand and simplify using the rules to find the resulting quaternion product.
Quaternion Division
Division of quaternions is not as straightforward as division of real numbers. It involves multiplication with an inverse. Here's how to proceed:
  • First, find the inverse of the divisor quaternion.
  • Multiply the dividend quaternion by this inverse.
To calculate \( \alpha / \beta \), where \( \alpha = 3 + 4i + 7j + k \) and \( \beta = 2 - 3i + j - k \), first find \( \beta^{-1} \), and then compute \( \alpha \cdot \beta^{-1} \). Simplifying the expression will yield the division result. This approach ensures the division respects quaternion properties.
Quaternion Inverse
To divide quaternions, you need the inverse of the divisor. The inverse of a quaternion is like a reciprocal in the sense of division. Here is the process:
  • Find the conjugate of the quaternion. The conjugate of \( \beta = 2 - 3i + j - k \) is \( \text{Conj}(\beta) = 2 + 3i - j + k \).
  • Calculate the norm squared: \( ||\beta||^2 = 2^2 + (-3)^2 + 1^2 + (-1)^2 = 15 \).
  • Now, use the formula for the inverse: \( \beta^{-1} = \frac{\text{Conj}(\beta)}{||\beta||^2} \).
With this inverse, you can perform division by multiplying the original quaternion by the inverse.
Complex Numbers Extension
Quaternions extend complex numbers by introducing three dimensions of imaginary numbers. This extension enriches the mathematical toolkit, allowing for calculations in four-dimensional space. While complex numbers use just one imaginary unit \( i \), quaternions use \( i, j, \) and \( k \).
  • Quaternion rules like \( i^2 = j^2 = k^2 = ijk = -1 \) provide new ways to approach problems.
  • They are extremely useful in three-dimensional computations like rotations in computer graphics.
By understanding the foundation of quaternions, you can tackle more complex scenarios than those solvable with just complex numbers. This makes them a powerful tool in advanced mathematics and physics.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Use Gauss's method to solve the cyclotomic equation \(x^{6}+\) \(x^{5}+x^{4}+x^{3}+x^{2}+x+1=0\)

Interpret the remaining three of Boole's equations \(x \bar{y} z=0\), \(x \bar{y} \bar{z}=0, \bar{x} y z=0\) from the case in the text, where \(x\) stands for clean beasts, \(y\) for beasts that divide the hoof, and \(z\) for beasts that chew the cud.

Show that in the domain of complex integers of the form \(a+b \sqrt{-5}\), the integers \(2,3,-2+\sqrt{-5},-2-\sqrt{-5}, 1+\) \(\sqrt{-5}, 1-\sqrt{-5}\) are all irreducible.

Show that the matrix $$ L=\left(\begin{array}{cc} \frac{a+Y}{X} & \frac{b}{X} \\ \frac{c}{X} & \frac{d+Y}{X} \end{array}\right) $$ where \(X=\sqrt{a+d+2 \sqrt{a d-b c}}\) and \(Y=\sqrt{a d-b c}\), is the square root of the matrix $$ M=\left(\begin{array}{ll} a & b \\ c & d \end{array}\right) $$

Design a lesson explaining negative numbers using either Peacock's principle of permanence of equivalent forms or Hamilton's formulation via pairs of positive numbers. Which formulation would work better in a classroom? Why?
See all solutions

Recommended explanations on Math Textbooks

Geometry

Read Explanation

Mechanics Maths

Read Explanation

Discrete Mathematics

Read Explanation

Theoretical and Mathematical Physics

Read Explanation

Pure Maths

Read Explanation

Logic and Functions

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.